Your space within VTEPS, Inc., which is poorly insulated, is expecting heat loss through the exterior walls that will cost the company $4,000 next year. A local company is offering insulation that can reduce the heat loss by 80%, with an installation cost of $17,000 now. Your team expects to work in the facility for the next 10 years. If it is estimated that the cost of the heat loss increases by $300 per year, after next year, using an MARR = 18%, find: PW, FW, AW, IRR
The firm is expecting a heat loss worth $4,000 next year. A new machine can reduce heat loss by 80%, with an installation cost of $17,000 now. So you pay 17,000 now, save 4,000*80% = $3200 next year and your saving increases by $300*80% = $240 per year, after next year. The new insulation will be operative for 10 years. There is a MARR = 18%.
First find the PW = -17000 + (3200(P/A, 18%, 10) + 240(P/G, 18%, 10))
= -17000 + 3200*4.4941 + 240*14.3525
= 825.72
FW = 825.72(F/P, 18%, 10) = 825.72*5.2338 = $4321.65
AW = 825.72(A/P, 18%, 10) = 825.72*0.2225 = $183.73
IRR = found by trial and error as 19.225%
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